Problem: Graph this system of equations and solve. $y = \dfrac{6}{5} x + 4$ $-6x-10y = 50$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$ $10$ $\llap{-}2$ $\llap{-}3$ $\llap{-}4$ $\llap{-}5$ $\llap{-}6$ $\llap{-}7$ $\llap{-}8$ $\llap{-}9$ $\llap{-}10$ Click and drag the points to move the lines.
The y-intercept for the first equation is $4$ , so the first line must pass through the point $(0, 4)$ The slope for the first equation is $\dfrac{6}{5}$ . Remember that the slope tells you rise over run. So in this case for every $6$ positions you move up You must also move $5$ positions to the right. $5$ positions to the right. $6$ positions up from $(0, 4)$ is $(5, 10)$ Graph the blue line so it passes through $(0, 4)$ and $(5, 10)$ Convert the second equation, $-6x-10y = 50$ , to slope-intercept form. $y = -\dfrac{3}{5} x - 5$ The y-intercept for the second equation is $-5$ , so the second line must pass through the point $(0, -5)$ The slope for the second equation is $-\dfrac{3}{5}$ . Remember that the slope tells you rise over run. So in this case for every $3$ positions you move down (because it's negative) You must also move $5$ positions to the right. $5$ positions to the right. $3$ positions down from $(0, -5)$ is $(5, -8)$ Graph the green line so it passes through $(0, -5)$ and $(5, -8)$ The solution is the point where the two lines intersect. The lines intersect at $(-5, -2)$.